Problem 12
Question
The angular elevation of tower \(C D\) at a point \(A\) due south of it is \(60^{\circ}\) and at a point \(B\) due west of \(A\), the elevation is \(30^{\circ}\). If \(A B=3 \mathrm{~km}\), the height of the tower is (A) \(2 \sqrt{3} \mathrm{~km}\) (B) \(2 \sqrt{6} \mathrm{~km}\) (C) \(\frac{3 \sqrt{3} \mathrm{~km}}{2}\) (D) \(\frac{3 \sqrt{6} \mathrm{~km}}{4}\)
Step-by-Step Solution
Verified Answer
The height of the tower is 3 km.
1Step 1: Set up the problem using trigonometry
Let's denote point D as the top of the tower and point C as the base of the tower. The problem provides two different angular elevations at points A and B. The distance from A to the base of the tower (point C) is due to the angular elevation of 60°, and will be denoted as AC. Using trigonometry, for point A (angle of elevation = 60°), the height of the tower CD is equal to \(AC \cdot \tan(60^\circ)\). Thus, \(CD = AC \cdot \sqrt{3}\).
2Step 2: Understand point B's alignment and use trigonometry
Point B is due west of point A and forms a right triangle with point C (the base of the tower) and point D (the top of the tower). The distance AB = 3 km. Using the right triangle ADC for the angle at B (30°), the length AC (adjacent to A) can be expressed in terms of the tower's height: \(CD = AC \cdot \tan(30^\circ) = AC \cdot \frac{1}{\sqrt{3}}\).
3Step 3: Equate expressions for the height from angles at A and B
By equating the two expressions for the height derived from the angles above, we have: \(AC \cdot \sqrt{3} = \frac{AB}{\sqrt{3}} \cdot \sqrt{3}\). Substitute the known value AB (3 km): \(AC \cdot \sqrt{3} = 3\). This simplifies to \(AC = \sqrt{3}\) km.
4Step 4: Calculate the tower's height
Now, using the value of AC obtained in Step 3 (\(AC = \sqrt{3}\) km), substitute it back into the height equation at point A: \(CD = AC \cdot \sqrt{3} = \sqrt{3} \cdot \sqrt{3} = 3\) km.
Key Concepts
Angular ElevationTowers and HeightsTrigonometric Identities
Angular Elevation
Angular elevation is a vital trigonometric concept, extensively used in solving various real-world problems, including determining the height of towers and buildings. When you are looking up at the top of an object from a certain point, the angle between the horizontal line from your eye and the line of sight to the top object is known as the angle of elevation.
The angular elevation provides crucial information about the slope or inclination you are observing. It often appears in problems involving measuring the height of distant objects that you cannot measure directly.
The angular elevation provides crucial information about the slope or inclination you are observing. It often appears in problems involving measuring the height of distant objects that you cannot measure directly.
- In our exercise, two different angles of elevation, 60° at point A and 30° at point B, are provided to calculate the height of the tower.
- Finding the height relies on these angles using trigonometry, making it a straightforward case of applying basic trigonometric identities.
Towers and Heights
In trigonometry, analyzing towers and heights is a common application, particularly when direct measurement is impractical. The challenge often involves determining the height of a vertical structure, like a tower, from points at a known distance from its base. To solve these problems, one usually forms right-angled triangles by considering the ground distance and the line from the observation point to the tower's top.
- In this exercise, the height of tower CD is calculated by forming right triangles using the points A and B.
- Point A provides a ground distance from the tower's base, while point B helps cross-verify using a different angle.
Trigonometric Identities
Trigonometric identities serve as fundamental tools for solving problems involving angles and distances. They are equations involving trigonometric functions that hold true for specific angles. These identities simplify the calculation processes, making it easier to find missing lengths or angles.
In scenarios like the one in our exercise, we make use of the tangent function, expressed as the ratio of opposite to adjacent sides in a right triangle.
In scenarios like the one in our exercise, we make use of the tangent function, expressed as the ratio of opposite to adjacent sides in a right triangle.
- At point A, where the elevation angle is 60°, the tangent formula provides that height = adjacent side × tan(60°), simplifying to height = adjacent side × √3.
- Similarly, at point B with an angle of 30°, the identity tan(30°) = height/adjacent side, simplifies to height = adjacent side × 1/√3.
Other exercises in this chapter
Problem 9
A person standing at the foot of a tower walks a distance \(3 a\) away from the tower and observes that the angle of elevation of the top of the tower is \(\alp
View solution Problem 10
The angle of elevation of the top of a tree at point \(B\) due south of it is \(60^{\circ}\) and at a point \(C\) due north of it is \(30^{\circ} . D\) is a poi
View solution Problem 13
An isosceles triangle of wood of base \(2 a\) and height \(h\) is placed with its base on the ground and vertex directly above. The triangle faces the sun whose
View solution Problem 14
The length of the shadow of a rod inclined at \(10^{\circ}\) to the vertical towards the sun is \(2.05\) metre when the elevation of the sun is \(38^{\circ} .\)
View solution